First of all, it is easiest to work in polar coordinates, r and θ, the robot starts at r=0 and moves along θ=θ

_{0}and you start at r=r

_{0}and θ=0 The robots path given

r(t) = v_{r}t

θ(t) = θ_{0}

We get to pick our r(t) and θ(t) and choose a path such that we will meet up with the robot path for any arbitrary value of θ

_{0}.

First of all, we might as well move straight to the robot, assuming that θ

_{0}is 0. We move toward r=0 and would meet the robot at r=r

_{0}*(v

_{r}/(v

_{p}+v

_{r})) This value, which I will call R, will be our effective starting point.

Next, we must move outward in a spiral, such that our r(t) continues to match the robot r(t), but our value of θ takes on all values between 0 and 2π. This is simple enough, find r(t) and θ(t) such that:

r'(t) = v_{r}

(r(t)θ'(t))^{2}+r'(t)^{2}= v_{p}^{2}

Solving it out, you get

r(t) = v_{r}t+R

θ(t) = √(v_{p}^{2}/v_{r}^{2}-1)ln((v_{r}t+R)/R)

Note the problem we would have if the robot were faster than the player. Anyway, since the logarithm is unbounded from above, you know that this θ will eventually reach 2π, solving the problem. This also finds the smallest disc you can solve the problem on, since you just find that t where θ is 2π and put it back into r.

Not too tricky of a problem once you hit on the solution really, but at first it did look somewhat impossible.

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